Given int arr[3][4][2]= {{{2,4},{7,8},{3,4},{5,6}},{{7,6},{3,4},{5,3},{2,3}},{{8.9},{7,2},{3,4},{5,1}}}; Which one of the following can be used to refer to element 1 in the above 3-d array?
*(*(*(arr+2)+3)+1)
In C, arr[i][j][k] is defined as *(*(*(arr+i)+j)+k): each subscript is pointer arithmetic on the array name followed by a dereference. Valid indices for int arr[3][4][2] are i = 0–2, j = 0–3, k = 0–1.
Step 1 — The initializer lists three 4×2 blocks. The third block (i = 2) is {{8.9},{7,2},{3,4},{5,1}}; its fourth row (j = 3) is {5,1}, and 1 is its second element (k = 1). So element 1 is arr[2][3][1].
Step 2 — Rewriting with pointers: arr+2 points to block 2, *(arr+2)+3 points to its row 3, *(*(arr+2)+3)+1 points to the element at k = 1, and the outer * reads it: *(*(*(arr+2)+3)+1) = arr[2][3][1] = 1.
Step 3 — arr[3][4][2] uses the declared sizes as indices and lies outside the array; arr[2][3][2] uses k = 2, past the last valid index 1 of the innermost dimension; *(arr+2)+*(arr+3)+*(arr+1) tries to add pointers to sub-arrays (and arr+3 is past the end), which is not an element access at all.
*(*(*(arr+2)+3)+1) option is correct.
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#include<stdio.h>
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